All of your math is correct except that h is just 36.7 because the bungee is attached to his feet not halfway up his body. Therefore his potential energy has to be measured to that point. Then you get
mgh=1/2kx^2
k=(2mgh)/x^2
k=(2*93.1*9.8*36.7)/9.3^2 = 774.29 N/m
6. A tall bald student (height 2.1 meters and mass 93.1 kg) decides to try bungee jumping from a bridge. The bridge is 36.7 meters above the river and the "bungee" is 25.3 meters long as measured from the attachment of the bridge to the foot of the jumper. Treat the "bungee" as an ideal spring and the student as a 2.1 meter rod with all the mass at the midpoint. This particular student desires to stay dry. What is the minimum spring constant (N/m) of the "bungee" that will allow the student to get as close as possible to the water but still stay dry? Assume that he begins at a standing position and "falls" from the bridge.
I made the energy formula:
mgh = 1/2kx^2
where h = height of bridge + half of student = 37.75
and x = height of bridge - length of student - length of bungee (to see how much it stretched) = 9.3m
BUT when I solve for k I don't get the right number, which should be 774 N/m.
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